Simple calculus question that is bugging me -answered,thanks
Posted: Mon Oct 21, 2013 4:34 pm
I'm sure this has a very "duh" answer, but it never occurred to me before and now it's really bothering me. Add it to the long list of things I never questioned about math until I started teaching it.
Consider the piece-wise function {x^2 if x<0; x^2 + 1 if x>=0}. The function is discontinuous at x=0, therefore the derivative does not exist. But the derivative is a limit, and a limit will exist as long as the right and left hand sides are equal. For this function, the left hand derivative is equal to the right hand derivative since both are zero, so doesn't this mean the limit and therefore the derivative exists? I'm trying to think of an argument based on real analysis, but it's been about 15 years since I took the class and I'm a little rusty. I need an explanation I can give to high school students.
Consider the piece-wise function {x^2 if x<0; x^2 + 1 if x>=0}. The function is discontinuous at x=0, therefore the derivative does not exist. But the derivative is a limit, and a limit will exist as long as the right and left hand sides are equal. For this function, the left hand derivative is equal to the right hand derivative since both are zero, so doesn't this mean the limit and therefore the derivative exists? I'm trying to think of an argument based on real analysis, but it's been about 15 years since I took the class and I'm a little rusty. I need an explanation I can give to high school students.